There are some assumptions in there that you haven't made explicit. They're correct, but they make the proof feel a little "cheaper".

1. Let K be the set of positive integers {n: n√2 ∈ ℤ}

2. Supposing K is non-empty, denote its minimum as k.

3. Consider k' = k√2 - k = k(√2 - 1)

Lemma 1: k' < k

L1. When x > 1, √x < x. Therefore, √2 < 2, √2 - 1 < 1, and k' = k(√2 - 1) < k.

Lemma 2: k' is a positive integer

L2a: By hypothesis, k√2 ∈ ℤ.

L2b: By hypothesis, k is a positive integer.

L2c: The difference of any two integers is itself an integer, so k' ∈ ℤ.

L2d: When x > 1, √x > 1. Therefore, √2 > 1, k√2 > k, and so k' = k√2 - k > 0.

4. k'√2 = (k√2 - k)√2 = 2k - k√2 ∈ ℤ. Therefore, k' ∈ K.

5. This contradicts the assumption that k is the minimum element of K. Thus K must be empty.

To make this proof work, we needed some general information about the behavior of the square root function over the entire real domain. (Or, some special knowledge about the numerical value of √2, but from an elegance perspective that's even worse.) This is something you can easily get away with, but it's somewhat annoying. The proof based on "assume this rational number is represented in its simplest form, such that gcd(n,d) = 1" seems a little more self-contained.